

DETROIT PUBLIC SCHOOLS 

Department of Instruction, 
Teacher Training and Research 



ELEMENTARY PROJECTION, DEVELOPMENTS 

AND 

SHEET METAL DRAFTING 




Published by the Authority of the 

Board of Education 

City of Detroit 

1921 



■-m 



These outlines for Elementary Pro- 
jection, Developments and Sheet Metal 
Drafting were prepared by Frank R. 
Kepler, Supervisor of Mechanical Draw- 
ing, Detroit Public Schools, and copy- 
righted by him in 1921 for the City of 
Detroit. 



DETROIT, PUBLIC SCHOOLS 

Department of instruction, 
Teacher Training and ResearchJ 



ELEMENTARY PROJECTION, DEVELOPMENTS 

AND 

SHEET METAL DRAFTING 




Pablished by the Authority of the 

Board of Education 

City of Detroit 

1921 



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^^'^-9/92/ ©CI.A6i4041 



ORTHOGRAPHIC PROJECTION 

Since all working drawings are made upon the principles 
of orthographic projection it is necessary before proceeding 
further to make a rather careful study of those principles. 
Orthographic projection simply means the drawing out on 
paper of forms showing their exact form and position with 
relation to co-ordinate planes. 

Perspective: — Looking at an object, visual rays extend 
from points of the object to the eye, all converging at the eye. 
If a transparent screen is placed between the object and the 
eye a picture is formed on the screen. This picture, or per- 
spective, never appears as the object actually is. Thus, by 
perspective, it is impossible to show proper proportions or 
exact dimensions. Fig. 1. 



P/CrU/^Z PLANE 
VJSUAL PAY^ 




r/6. /. PEPSP^C T/VE 

Orthographic Projection: — In perspective the point of 
sight is always at a given or finite distance from the object, 
while in orthographic projection the point of sight is consid- 
ered at an infinite distance and the object at a finite distance 
from the screen, or plane. Hence the visual rays are consid- 
ered perpendicular to the plane, or screen, and parallel to each 



other. Consequently all forms appear in their exact relation 
to the plane. Fig. 2. 



PL A /VE or P/^OJECT/ON 

PRO^J£:C^/NG L/A/E 




Ci 



3^ 




f/^OJ£CT/ON 



no. 2.. O/^THOG/^/^PH/C PROJECTION 

Terms Used:^In order to discuss the principles of ortho- 
graphic projection it will be necessary to define certain terms 
used in the discussion. Let a point, as B, be assumed in space. 
Fig. 3. A straight line drawn from the point perpendicular to 
a plane is the projecting line, B W. The plane to which it is 
drawn is the plane of projection. The point where the. line 
meets, or pierces, the plane is the projection of the point 
assumed, b'. Thus it will be seen that a projecting line is 
nothing more or less than a visual ray from the point assumed 
to the point of sight, at an infinite distance and the projection 
is the piercing point of the visual ray. 



Planes of Projection: — A horizontal plane of projection and 
a vertical plane of projection are assumed in orthographic pro- 
jection. These are mutually perpendicular. Projecting lines 
are drawn perpendicular to these planes of projection. Hence 
projecting lines are either horizontal or vertical. The 
projection of an object on the vertical plane oi projection is 
its vertical projection, b', h', k', c', Fig. 3. The projection on 
the horizontal plane is its horizontal projection a, b, c, d. Pro- 
jections are named from the planes on which they are made. 



A plane of projection perpendicular to both the horizontal 
and vertical planes is often required. This plane is called the 
profile plane of projection. A projection made upon this plane 
is called a profile projection, a'^ f , h", b'', Fig. 3. 




An auxiliary plane is frequently used to give a view of the 
exact shape and size of a surface that cannot be shown on the 
horizontal, vertical, or profile planes. 

Views: — View in connection with working drawings means 
a projection. The horizontal projection of an object is often 
called the Top view or Plan; the vertical projection is called a 
Front view or Elevation; and the profile projection is called an 
End view, or an End Elevation. 



Projection of Lines, Surfaces, and Solids: — Since a line is 
determined by its points, its projection is determined by the 
projection of these points on the planes of projection. When 
the extremities of a straight line have been projected onto a 
plane of projection, a straight line joining these projections is 
the projection of the line. Observe in Fig. 3 that the projec- 



tions of the points B and C at b and c give the extremities of 
the horizontal projection of the hne BC. 

Since a surface is bounded or determined by lines, the 
projections of these bounding lines determine the projection of 
the surface. 

A solid is determined by bounding surfaces, and its projec- 
tions by the projections of the bounding surfaces. 

Therefore to find the projection of a solid, first find the 
projection of the points determining the lines which bound the 
determining surfaces of the solid, then connect these with the 
proper lines. 

Revolving the Planes: — A drawing is made in one plane, 
the plane of the paper. It is therefore necessary to revolve 
that portion of the H plane behind the vertical about its inter- 
section with the V plane until it becomes above and in the 
same plane with the V plane, Fig. 4. The intersection of the 



d 




q_ 


— -^^ 


^ 






1 % ^-^ 






\ \ \ 

Jb' "~-^ \ v^ \ 


c 




c' 


</' 


jb'^ 


■ a' 


x3' 


■ C" I \ct'-d 




e' 








1 1 


A- 


J- 


f 


s/- 


A- f:e" 



H plane and the V plane is called the ground line (G. L.) On 
the drawing the ground hne is represented by a horizontal 
line. All that i)ortion of the paper above or behind the G. L. 
represents tlie H i)lane while all that i)ortiou below the G. L. 
represents the V plane. 



6 



In the same manner, the Profile plane is revolved about 
its intersection with the V plane until it is in the same plane 
with the V plane. The intersection of the Profile plane and 
the vertical plane is called the profile line (P. L.) On the 
drawing the P. L. is represented by a vertical line and there- 
fore perpendicular to G. L. Since the P plane is placed either 
to the right or the left of the object according to the view or 
projection required, the P plane may be revolved to the right 
or left, as the case may be. 

Fig. 3 illustrates a piece, ABCDEFJK, located in the 
third angle below H and behind V with the Profile plane at 
the right. The horizontal projection of the grooved block is 
shown at a b c d, the vertical projection at o' b' j* k^ and the 
profile projection at a'^ V y W\ When the H plane, M L' L G 
has been revolved about G. L. and the V plane, L L' N P has 
been revolved to the right about P. L., the H, V, and P planes 
all lie in the same plane. This may be taken as the plane of 
the paper. Fig. 4 illustrates the position of the projections 
when they have been brought into the same plane. Observe 
that the H and V projections of any given point on the piece 
lie in the same vertical line, and that the V and the P projec- 
tions lie in the same horizontal line. 

Third Angle: — The horizontal and vertical planes (consid- 
ered infinite in extent) are perpendicular to each other and 
form four right dihedral angles. Fig. 5. Let the horizontal 
plane be denoted by H and the vertical plane by V. An object 
lies in the 



r/RSr AAfGL£ 



SECOND A A/GLE. 




First angle when it is above H and in front of V. 
Second angle when it is above H and behind V. 
Third angle when it is below H and behind V. 
Fourth angle when it is below H and in front of V. 

7 



It is almost the universal practice in this country to use the 
third angle since it affords a more convenient arrangement of 
the views or projections for working drawings and is more 
modern. The first angle is still used to some extent in this 
country and almost universally in European countries, 
although even there the use of the third angle is slowly becom- 
ing more common. 

The third angle unless otherwise noted is used in this 
discussion and in the problems for practice. 

Relative 'Position of Views — Any point in the third angle 
has its H projection above the ground line, its V projection 
below the Ground Line in the same vertical line with the H 
projection. It may further be stated that the profile projec- 
tion and the vertical projection are in the same horizontal 
line. 



^HORIZONT^L PL.^ 



Of' y^ 

VELRTICAL. /=>£. 



^f(,V 



riG. e 



H 
T a 


''"■ V 



r/G. 7 



In Pig. 6, if A is a point in space then Aa and Aa' are pro- 
jecting lines which determine the H projection and the V 
1 rojection, respectively. Now, if a plane is passed through 
the point A and also through the projecting fines Aa and Aa', 
il will intersect H on the line ab and V on the fine a'b, both of 
which are perpendicular to the G. L. When H and V have 
been revolved into the same plane ab and a'b will lie in the 
same straight line perpendicular to the Ground Line. Fig. 7. 



8 



In drawing, when one projection has been determined, the 
other may be determined by drawing a straight Une perpen- 
dicular to the G. L. through the projection first found and 
locating the other projection on this line the proper distance 
from the G. L. The hne so drawn is a Line of Projection and 
should not be confused with a Projecting Line. A line of pro- 
jection exists only on the drawing; a projecting line exists 
only in imagination. A line of projection might be said to be 
a projection of a projecting line. Observe in Fig. 3, that B b 
is a projecting line, and that, in Pig. 4, b b' is a line of pro- 
jection. 

A draftsman must first be able to conceive each view or 
projection on its corresponding plane of projection and then 
determine the positions of the different views with respect to 
each other and to the Ground Line when the planes of projec- 
tion have been brought into the same plane. This requires 
considerable practice and should become so natural that one 
should not be conscious of the first step at all. 

PRINCIPLES OF PROJECTION 
A Point in Space 

(1) Every point has two views, one on H, the other on V, 
by which its position in space is determined. Fig. 6. 

(2) A point and its two views he in the same plane per- 
pendicular to both H and V — ^i.e., a plane passed through the 
two projecting lines. Fig. 6. 

Since the projecting line, Aa in Fig. 6, is perpendicular to 
H a plane passed through Aa is perpendicular to H. Simi- 
larly Aa' is perpendicular to V. Therefore a plane passed 
through the point A and its two projecting lines is perpendicu- 
lar to both H and V. This plane is called a Projecting plane. 

(3) The Top view of a point is as far behind the Ground 
Line as the point itself is behind V. The Front view of a point 
is as far below the Ground Line as the point itself is below H. 
Fig. 4. 

(4) The two views of a point always he in the same 
straight line, perpendicular to the Ground Line. Fig. 7. 

The intersections, ab, with H, of the plane passed through 
Aa is perpendicular to V and to the Ground Line at b and the 



intersection, a'b with V of the plane passed through Aa is 
perpendicular to H and the Ground Line at b. Therefore when 
the horizontal plane (H) has been revolved until it is in the 
same plane with V, ab and a'b will lie in the same straight line 
perpendicular to the Ground Line. Both ab and a'b are per- 
pendicular to the Ground Line and have the point b in com- 
mon. 

A Line in Space 

(5) A line perpendicular to either plane of projection has 
for its view on that plane simply a point, Fig. 8. 

(6) A line perpendicular to either plane of projection has 
for its view on the other plane a straight line perpendicular to 
the Ground Line and equal in length to the line of which it is 
the projection. Pig. 8. 

It is evident that the intersection of the plane of projec- 
tion and the projecting plane of the line is perpendicular to 
the Ground Line. 





r/G. 3 



(7) When a line is parallel to either plane of projection 
its view on that plane represents the true length of the line; 
and the angle which this view makes with the Ground Line is 
equal to the angle which the line in space makes with the 
plane to which it is not parallel, Fig. 9. 

(8) A line parallel to either plane of projection has for 
its view on the other plane a line parallel to the Ground Line, 
Fig. 9. 



10 



(9) When a line is not parallel to a plane of projection, 
its view on that plane is always shorter than the true length 
of the Une, Fig. 9. 

(10) A Une parallel to both H and V has for its two 
views, lines parallel to the Ground Line, both of which are 
equal in length to the line itself. 

(11) If a hne is parallel to neither plane of projection, 
both views are shorter than the line itself. The angles which 
the line makes with the planes of projection are not repre- 
sented in their true size by the angles which the views make 
with the Ground Line, Fig. 10. 




In general, the projection of a line parallel to a plane of 
projection is equal to the line itself. And, the projection of a 
line at an angle with a plane of projection is shorter than the 
line itself. A projection of a line is either equal to or shorter 
than the line itself. Since surfaces are bounded by lines, this 
principle is also true of surfaces. 

In order to find the true size and shape of either a line or 
surface not parallel to the H, V, or P plane of projection, it is 
necessary to either revolve the line or surface until it is paral- 
lel with one of these planes ; or assume a plane parallel to the 
surface or line. A plane so assumed is an auxiliary plane of 
projection. 



11 



AUXILIARY VIEW 



It is often necessary to make a view of an object that is in 
such a position that a suitable view is not shown on the hori- 
zontal, vertical, nor profile planes. As in the case of the 
truncated square prism, Pig. 11, the true size and shape of the 
section cut by the plane is not shown either on the horizontal, 




vertical, or profile plane. A plane, parallel to the cutting 
plane, is used and the projection is made on this plane. A 
plane so used, or assumed, is called an auxiliary plane of pro- 
jection. The projection made on this plane, of the section 
made by the cutting plane, will be the true size and shape of 
the surface, for the surface is parallel to the auxiliary plane. 

12 



It will be observed in Fig. 11, that the Trace of the Auxil- 
iary Plane on V is parallel to the Trace of the Cutting Plane 
on V, for the auxiliary plane is assumed parallel to the cutting 
plane. Projecting lines drawn from the object to the auxiliary 
plane of projection are perpendicular to this plane. The 
traces of a projecting plane, passed through one of these pro- 
jecting lines perpendicular to the V plane, will form lines of 




riG. /& 
projection on the auxihary plane and on the V plane. When 
the auxihary plane has been brought into the same plane with 
H and V, the corresponding lines of projection on V and on the 
auxihary plane wih he in the same straight hue perpendicular 
to the auxiliary trace on V and also to the trace of the cutting 
plane on V, as shown in Fig. 12. 



13 



First Method: — Fig. 12. In the drawing, make the Trace 
of the auxiliary plane on V parallel to the Trace of the cutting 
plane and also locate the auxiliary trace on H before the planes 
have been revolved and also when the auxiliary plane has 
been revolved into the same plane with V. The trace on 
H will now be perpendicular to the auxiliary trace on V, and 
is indicated as the trace of auxiliary plane on H', Fig. 12. 
Draw the lines of projection to the auxiliary trace on H and 
with the intersection of this trace with ground line as a center 
swing the lines of projection to the trace of auxiliary plane on 
H'. Continue the lines of projection parallel to the auxiliary 
trace on V. Draw the lines of projection from the V projec- 
tion perpendicular to the auxiliary trace on V. The auxiliary 
view is determined by the intersection of the corresponding 
lines of projection. 

Second Method: — Fig. 12. Another method, which is often 
more convenient, is employed. In this case an axis is drawn 
through the section made by the cutting plane, the H projec- 
tion of which is represented by the line AB, Fig. 12. Locate 
this axis on the auxiliary plane by the hne A'B' parallel to 
and at a convenient distance from the trace of the cutting 
plane. Draw the lines of projection from the V projection 
perpendicular to A'B'. With the dividers, measure the dis- 
tances on the corresponding lines of projection from the line 
A'B' on the auxiliary plane. Connect the points thus found 
and determine the auxiliary view. 




b 


o 




9 


1 
\0 


ni' \ 


\ 


1 

1 1 






nL.'^. 


Y-'k. 



n 




C 



h' cf' 



t>" a" c" ct 



rfG. t3 
Third Method: — -Fig. 13. Surface revolved into position 
parallel to H. This method is very convenient and may be 



14 



used to great advantage. If the surface bounded by HE, EF, 
FG, and GH be revolved about G until it is parallel to the H 
plane then the projections will assume the form shown in 
Fig. 13. With g as center, revolve ge to the horizontal posi- 
tion, as gn'; the points f and h will fall at o' and m'. Lines of 
projection drawn from the points m', n', and o', to the cor- 
responding lines drawn from d, a, and b give the horizontal 
projection of F, H, and E. Then the parallelogram bounded 
by the lines cm, mn, no, oc is the horizontal projection of the 
surface after it has been revolved into a position parallel to H. 
It is more convenient however to place this view to one side, 
as follows: 

Simply project horizontal lines from the points a, b, c, and 
d. Locate g'' at a convenient position on cz, set off g''x equal 
to gf, xy equal to fh, and yz equal to he. At x draw xf' per- 
pendicular to g''z; at y, the perpendicular yh'^ and at z, the 
perpendicular ze'^ The intersection of the line bf ' and xf 
is the location of the point f'; h'' and e'' are found in the same 
manner. 

True Length of Line: — In Fig. 14, let AB be a line parallel 
to neither H nor V. Then ab and a'b' are its projections on 
the horizontal and vertical planes respectively. If the point 
A remain fixed and the line be revolved until it is parallel to 
V the point B will move to M, in an arc, the plane of which is 
parallel to H. The projection of this arc on H is bm and on V 
is the Une b'm'. The horizontal projection of AB in its new 
position is am parallel to G. L. for the line AB is now parallel 
to V. The V projection of AB in its new position is a'm', and 
is the True Length of AB, for it has been made parallel to V. 





^' ^ 



J^/G. /S 



On the drawing. Fig. 15, using a as center and radius ab, 
draw the arc bm and the straight line, am, parallel to G. L. 
Draw b'm' parallel to G. L. on V. Draw the vertical Une mm' 



15 



intersecting b'm' at m'. Connect a' and m'. Then a'm' is the 
True Length of the hne. Since the extremity of the hne moves 
in a plane parallel to the plane of projection, any point in the 
line will move in a plane parallel to this plane and to the plane 
of projection. 

A method used to great advantage in practical work is as 
follows: In Fig. 16, Ab is the H projection of AB. Bb, the 
projecting line of B, is perpendicular to the H plane and there- 
fore to Ab. Then AbB forms a right-angled triangle of which 
AB is the hypotenuse, Bb the altitude, and Ab one leg. When 
the H plane has been revolved into the same plane, Fig. 17, ab 




r/G. /6 




rfG. ii 

is the H projection, a'b' is the V projection, and mb' is the alti- 
tude of the line AB and of the V projection. In the right tri- 
angle m'on, Fig. 17, make m'n equal b'm and no equal ab, 
then m'o is the true length of the line AB. Since the triangles 
m'on, Fig. 17, and AbB, Fig. 16, are equal by construction, the 
hypotenuse of one is equal to the hypotenuse of the other. 

DEVELOPMENTS 

The ability to develop, or lay out, the surface of solids or 
forms is an important part of a draftsman's equipment. The 
laying out of such work is a large part of the sheet-metal 
worker's business. Many of the so-called practical methods 
or short-cuts employed by the trade may easily be explained 
by the principles of orthographic projection. A thorough 
understanding of these principles will make the practical ap- 
plications comparatively simple. 



16 



To lay out the surface for a truncated prism: — In Fig. 18, 
since all the edges of the prism are perpendicular to the base, 
straight lines representing the edges are drawn at right angles 
to a straight line representing the distance around the prism 
at the base. To find the points at which to draw these per- 
pendiculars, set off a''b'' equal to ab; b^'c'' equal to be; c"d'' 
equal to cd; and d^'a'^ equal to da. On these perpendiculars 



0) 






<b 


^ 




A 


V 




; 
O 


)/ 


1 \ 






5^ 






\ 


'^ 




\ 


(J 





0) 




set off the heights of each edge as a''e' equal to a'e; b'^h' equal 
to b'h; etc. Joining the points e'h', h'g', etc., the sides are 
laid out. The bottom and top may be joined in the proper 



17 



position. Care must be taken to place these in such a posi- 
tion that when folded into place the edges will meet their 
respective sides. Those surfaces must be of the true size and 
shape. 



To lay out the surface of a pyramid: — It will be observed 
that the lateral surfaces of a pyramid are triangular. If the 
edges bounding these surfaces are determined, the triangles 
forming the surfaces may be constructed. From an exam- 




riG. /-^ 



ination of Fig. 19, it will be observed that the triangles are 
adjacent with a common point at the vertex, one side in com- 
mon, and that the bases are not in a straight line. Observe 
that the length of the sides of the triangles is the true length 
of the edges of the pyramid in each case. 

In Fig. 20, are shown the projections of a truncated square 
pyramid. In order to find the development of its surface it is 
necessary to determine the true length of the edges. The 
true length of 0-2 is found to be 0'-2^, 2^ is found at the inter- 
section of a horizontal line drawn through 2^ and 0'-2i. 6^ is 
found at the intersection of a horizontal line drawn through 
6'. (See true length of line, Fig. 15) . 

Since all the edges in this right pyramid are equal, the arc 
2,-2, is drawn with 0' as center and 0'-2, as radius. With the 
chord 1-2 equal to one side of the base, set off the distances 
2i-1i, 1i-4i, 4i-3i and 3i-2i. Set off the distance O'-S, on the line 
0'-2, in the development equal to 0'-6,; 0'-1,, 0^-8^ equal to 
0'-7,. Other points are found in a similar manner. Complete 
the lay-out by joining the points and adding the truncated sur- 
face as well as the base. 



18 




3--f 



CURVED SURFACES 



A surface may be generated, or produced by moving a 
line. If the moving line be made to meet certain conditions 
or follow certain lines, definite surfaces may be generated. 



v/ / /A*/ / / / 



/ 



/.^/ 



if / />/ / / / 



w / 



/<// / / / 



^/ / /' /' / / / 

FIG. 21 

A straight line moving so as to touch two parallel lines would 
produce a plane surface. The moving line is the generating 
line. The Une giving the direction to the moving line is the 
directrix, the various successive positions of the generating 
line are the elements, Fig. 21. 



19 



Cylinder 

Every cylinder may be generated by moving a straight, 
or right hne, so as to touch a closed curve and have all its 
positions parallel. If a straight line be moved so that it 
touches a circle and its various positions are parallel to each 
other and perpendicular to the plane of the circle, a right 
circular cylinder will be generated. The moving line is the 
generating line; the various positions of the moving line form 
elements on the curved surface of the cylinder. 

The consideration of these elements on the curved surface 
of the cylinder is essential to the development of the surface, 
especially that of a truncated cylinder or of intersecting sur- 
faces. In the drawing, to locate any point on the curved sur- 
face, it is necessary to locate the element passing through 
this point. The element being located on any one of the 
different projections, it is easy to locate this point on the 
given element in any one or all of the views. 

If any point as A, Fig. 22, be located on the surface of a 
cylinder the vertical projection of which is a' a straight line 
drawn through a' parallel to the axis, will represent its ele- 
ment. The point a will not only be the horizontal projection 
of the point A but also of its element. The profile projection 
of the element may be made and thfe location of a'' may be 
projected from a'. 




\a1 




r/G. a a 



r/G. a 3 



To determine elements on the surface of a cylinder: — A 
plane passed through a cylinder so as to contain the axis will 
intersect the surface in two straight lines parallel to the axis. 
Since these two lines are parallel to the axis and therefore to 
each other, they are right line elements on the curved surface. 
Fig. 22. Again a plane may be passed through a cylinder 



20 



parallel to the axis and its intersection with the curved sur- 
face will be straight lines parallel to the surface and to each 
other. These lines, then, are right line elements of the curved 
surface, Fig. 23. 




p*-^ -.^ 


^> 


i.Liw:::: 


5' 


fcV. 


V 


L^ii"-: 


01' 


•v:!.: ^ 


<<>' 




Right line elements, then, may be found on the surface of 
a cyUnder in two ways: (1) By passing a plane through the 
cylinder parallel to the axis; or, (2), By passing a plane 
through the axis in any direction so that the axis is contained 
in the plane. Fig. 24 readily illustrates the method of laying 
out the surface of a cylinder. How long must the line I1-I1 
be? Why must the distances 1i-2^, 2^-3^, etc., on the develop- 
ment, be taken from the circumference in the H projection? 
Knowing the diameter, what is the formula for figuring the 
circumference? 



21 



Cone 

Every cone may be generated by moving a right, or 
straight, hne so as to touch continually a closed curve and 
pass through a fixed point not in the plane of the curve. 

The moving line is the generating line; the fixed point the 
vertex of the cone; and the different positions of the generat- 
ing line are the right-line elements. 

These right-line elements may be found on the surface of 
a cone by passing a plane either through its vertex in any 
direction so as to contain the axis of the cone, or through the 
vertex and the base of the cone in any direction. 

Elements thus found may be shown on the drawing by 
dividing the circumference of the base into a number of parts, 
preferably equal parts, and drawing lines from these points 
through the vertex as in Fig. 25. These lines represent the 




elements in the H projection. Project the points 2, 3, 4, 5, 
etc., to the base in the vertical projection as 2', 3', 5% etc. 
Connect the points thus found with 0', the apex. 

If a' in Fig. 26, is the V projection of a point on one of the 
elements as 0'-3', then its H projection may be found by 
projecting its position from a' to its position on 0-3 at a. 
Again if any point as P has its V projection at p', Fig. 25, a 
straight line representing an element is drawn from the apex 
through p' to the base. A projection of this element is then 
made on the horizontal projection and the point p found by 
projecting from p'. 



22 



To develop a right circular cone: — It is evident that in a 
right circular cone all the elements are of equal length. The 
line 0^-1 ^ is the true length of one of these elements, Fig. 26. 
Why? With a radius 0'-1^, draw an arc setting off distances 
1^-2^, 2i-3i equal to the divisions on the circumference in the 
H projection. Why? These points may be connected with 
the center of the arc representing the apex. The length of 
the arc l^-li must equal the circumference of the base. 




Since all the elements are equal, the true length of the 
element 0-3 coincides with O'-l^, and the true length of the 
distance 0-a will be shown at O'-ai. Why? See the pyramid, 
Pig. 20. The true distance from the apex to any point on any 
of the other elements may be found in the same way. 



23 



INTERSECTIONS 

At this stage, the notion should be firmly fixed that a sur- 
face is made up of elements. When two surfaces intersect, 
■ a point on the line of intersection is the point of intersection 
of an element of one surface with that of the second surface. 
These elements are determined by the passing of a plane 
through the solids in such a way as to determine elements on 
the surfaces of each. It is evident that an element 'cut out' 
by a plane on one surface will intersect an element 'cut out' 
by the same plane on the other surface; since, in a plane, all 
straight lines not parallel to each other intersect. The point 
of intersection of the two elements is a point common to 
both surfaces and is a point on the line of intersection. 

The method of finding the intersection of surfaces must 
be familiar to the well-equipped draftsman, as intersections 
are constantly occurring both in working drawings, and in 
sheet me^al work. In sheet metal work enough points must 
be founa to lay out the development accurately. 

Intersection of Cylinders 

If a plane be passed through the two intersecting cylinders. 
Fig. 27, parallel to the axis of each, the element AB will be 
determined on the large cylinder G, and also the elements CD 
and EF on the smaller, H. It will be noted that CD and EF 
intersect AB at D and F, points on the line of intersection. On 
the working drawing the plane M M 'cuts out' the element rep- 
resented by c" b'' on the smaller cylinder in the profile view 
and a' b' on the larger cylinder in the front view. 

The element CD appears in the front view at c' d', obtained 
by projecting from the top view. The intersection of c' d' and 
a' b' at d^ is a point on the line of intersection of the two 
cylinders; the point V is obtained similarly. Other points are 
found in like manner. 

It is more convenient to divide the circumference of the 
small cylinder into equal parts and so pass the planes as to 
determine elements equi-distant apart on the smaller cylinder. 
Time may be saved by drawing a semi-circumference on the 
H and the profile projections. 

In laying out the development of the surface of the small 
cylinder, it is necessary to set off along the straight line, 5,-5, 
distances equal to the length of the arc 1-2. As many dis- 

24 



tances must be set off as the whole circumference of the cyl- 
inder has been divided into, making sure to make the 'lay- 
out' check with the calculated circumference. At these points 
erect perpendiculars representing the elements. On these 
elements set off distances equal to the length of the elements, 
as obtained from either the horizontal or profile projection 
of the cylinder H. 




CD W 



^ 




o 


/ 




s 


Vj 


^ 


t 




In laying out the larger cyhnder, indicating the center of 
the opening, at 1 ^^ , erect a perpendicular to the straight hne rs 
representing the circumference of the cylinder. From l^^ set 



25 




off on each side along the distance 1ii-2ii equal to the arc 
1/'-2'^ the distance 2ii-dii equal to the arc 2'^-d'^ continuing 
thus set off the remaining distances equal to the corresponding 
arcs, being sure that the whole distance from r is equal to 
one-half the figured circumference of the cylinder. From the 
vertical projection is obtained the distances on the elements 
as is indicated. 



26 



Cylinders at an angle: — The intersection of two cylinders 
at an angle is found in a similar manner to that of those at a 
right angle. In Fig. 28, let mn, in the horizontal projection, 
indicate the trace of the plane MN passed through the two 
cylinders parallel to the axis of each. The elements 'cut out' 
on the oblique cylinder are shown in the H projection by rt 
and St; in the V projection by r'r' and s's'. The element 'cut 
out' on the vertical cylinder is shown in the V projection by t't'. 
The points at which r'r' and s's' intersect t't' as o and o' are 
points on the intersection of the two cyhnders. Other points 
on the intersection may be found in a similar manner. 

By means of the auxiliary semi-circle in the H projection 
the planes for 'cutting out' elements determine these at equal 
distances on the surface of the obhque cylinder. A similar 
semi-circle in the V projection serves to locate the position of 
the corresponding elements in that projection. 

In laying out the oblique cylinder, draw AB perpendicular 
to the axis. Let A'B', in the development, represent this Une. 
Along A'B' set off equal distances between the elements as 
determined from the auxiUary semi-circle; draw the elements 
perpendicular to A'B'; and set off from A'B', along the ele- 
ments, distances equal to the corresponding distance from AB 
in the vertical projection. 



27 



PROBLEMS 

Show the Horizontal and Vertical Projections 

Prob. 1. (a) A 21/4'' square card is parallel to H with one 
edge parallel to V. Front edge is y^' behind V and %" be- 
low H. 

(b) The same card, still having the front edge 
parallel to V, is turned downward on the right 30°. 

Prob. 2. (a) A 1%'' by 2^4'' rectangular card is parallel 
to V with the longer edge parallel to H. The card is %'' be- 
hind V; its upper edge y^' below H. 

(b) The same card, the lower edge remaining 
parallel to H, is turned backward on the right 30°. 

Prob. 3. (a) An equilateral triangular card is parallel to 
V with its lower edge parallel to H. Edges 2''. It is %'' be- 
hind V and the lower edge 2i/4'' below H. 

(b) The same card, except that the lower edge 
is turned downward on the right 15° to H. 

Prob. 4. (a) A 1%'' by 2'' rectangular card is parallel to 
V. The long edges make an angle of 30° with H. The card 
is 1/2'' behind V and its nearest corner %'' below H. 

(b) The same card, except that it is turned 
back on the right 30° to V. 

Prob. 5. (a) A 2'' square card is parallel to H. The 
edges make an angle of 45° with V. It is 1/2'' below H and its 
nearest corner 14 '' behind V. 

(b) The same card, except that it is turned 
down on the right 30° to H. 

Prob. 6. (a) An hexagonal plinth, %'' thick, is parallel 
to V with the upper edge parallel to H. Its sides are 1^/4 ''. It 
is %'' behind V and its upper edge 1/2'' below H. 

(b) The same plinth turned backward 45° on 
the right. 

Prob. 7. (a) A 1%'' square plinth, %'' thick, is parallel 
to H with the right hand edge 60° to V. The upper surface 
of the plinth is I/2'' below H; the nearest comer %'' behind V. 
(b) The same plinth in the same position, ex- 
cept that it is turned downward on the left 30° to H. 

Prob. 8. (a) An octagonal plinth, 1/2'' thick, is parallel 
to H, the extreme left-hand edge is 75° to V. This plinth is 3" 
across from side to side, is V below H, and the nearest corner 
of the enclosing square is 1/2'' behind V. 

(b) The same pinth, except that it is turned 
downward on the right 30°. 

Prob. 9. (a) A circular plinth, i^'' thick and 23/4'' i,i di- 
ameter^ the center of which is 1%'' behind V and %" below 

28 



H, is parallel to H. 

(b) The same plinth turned down on the left 
30° to H. 

In addition to the H and V projections show the Profile 
projections in the next six (6) problems. See page 7. 

Prob. 10. A triangular prism, 3%'' high and sides 1" , 3", 
and 3%'^ respectively, has the broadest side 75° to the right, 
the nearest corner y^' behind V. The base is ^" below and 
parallel to H. 

Prob. 11. An octagonal prism with 1%^' sides and 3%'^ 
high rests on an octagonal base parallel to H. One of the 
sides is 30° on the right with V. The nearest corner is %'' 
behind V and the lower base is 4'' below H. 

Prob. 12. A pentagonal pyramid, 3i/4'^ high and the base 
inscribed in a 3'' circle, has its base parallel to H and the 
nearer side of base parallel to V. The base is M' below H, the 
center of which is 2" behind V. 

Prob. 13. A triangular prism, 3'' long, with sides 4", 
2y2'^ and 4'^ rests on the 21/2'' hy V side with the bases 
parallel to P. The upper edge is parallel to and 21/2'' behind 
V; and parallel to and %^^ below H. 

Prob. 14. An equiangular triangular pyramid with 1%^' 
sides and Zy^'' high, has its base parallel to V and the upper 
edge of its base 15° to H. Base is 4'' behind V and upper cor- 
ner is 11/4'' below H. 

Prob. 15. An hexagonal pyramid 3%^' high with 1%'' 
sides at the base, has its base parallel to V and one side of 
base parallel to H. The base is %'^ behind V and nearest side 
3/4'' below H. 

Revolution of Objects 

Prob. 16. (a) An equilateral triangular plinth, 2'-0'' sides 
and l'-6'' thick, rests on a 2' x l'-6'' rectangular base parallel 
to and 3'-0' below H. The nearer triangular face is parallel 
to and I'-O'' back of V. Scale, l^''--l'-0''. 

(b) The same plinth in the same position, ex- 
cept that it is turned up on the left 45° to H. 

(c) The same plinth in the same position, ex- 
cept that it is turned back on the right 30° to V. 

Prob. 17. (a) A square pyramid, l'-6'' square at the base 
and 2'-6'' in altitude, rests on its square base parallel to and 
3'-0'' below H so that the sides of the base are at 45° with V, 
the nearest corner being 6'' back of V. Scale, li^''i=l'-0''. 

(b) The same pyramid in the same position 
except that its axis is inchned to the right 60° to H. 

(c) The same pyramid in the same position 
except that its axis is turned backward on the right 15° to V. 

29 



Problems in Development of Surfaces 

Show H, V, and P, projections, auxiliary view, and de- 
velopment in the following problems. See pages 12, 13, 14. 

Prob. 18. Truncated Square Prism. A square prism, with 
base parallel to H and the nearer side 30° to V, is cut by a 
plane 45° to H and perpendicular to V so that the prism is 3'^ 
high on the left. The base is 1%^^ square. 

Prob. 19. Truncated Triangular Prism. A triangular 
prism, with 2%'\ l>^^^ and 2%'' sides and the base parallel 
to H, has the broadest side at the rear and 15° to V. The 
prism is cut by a plane, 30° to H and perpendicular to V, so 
that it is 2%'' high on the extreme left. 

Prob. 20. Truncated Hexagonal Prism. The hexagonal 
base is parallel to H and is 2'' across the flats. The prism is 
cut by a plane, 30° to H and perpendicular to V, leaving it 
2%^' high on the extreme left. 

Prob. 21. Make the lay-out or 'stretch-out,' as it is 
termed in this case, of the mitred sheet metal moulding, the 
V projection of which is shown in Fig. 31. Scale, full size. 
Use only vertical projection and the bottom view. 




F/G. 31 



Prob. 22. Truncated Square Pyramid. A square pyra- 
mid, with base parallel to H and sides of base 45° to V, is cut 
by a plane at 45° to H and perpendicular to V, V2'' above the 
base on the right-hand edge. The base is 1%" square and 
the height of the original pyramid 3^''. 

Prob. 23. Truncated Hexagonal Pyramid. An hexagonal 
pyramid, with the base parallel to H and the nearest side of 
the base parallel to V, is cut by a plane, 45° to H and perpen- 
dicular to V, %'' above the base on the right-hand edge. The 
base is 2y^" across the corners; the original altitude 31/2''. 



30 



Prob. 24. Truncated Oblique Square Pyramid. The base 
is parallel to H, its side 45° to V. the axis is parallel to V but 
inclined to the right so that the apex is 21/2'' to the right of 
the center of the base. The cutting plane, perpendicular to 
V, makes an angle of 45° with H an<i cuts the extreme right- 
hand edge %^^ above the base. The sides of the base are 1%'^ 
the altitude of the original pyramid 31/2''. See page 15. 

Prob. 25. Truncated Oblique Pentagonal Pyramid. The 
base is parallel to H with the right side of base perpendicular 
to V. The extreme left-hand edge is parallel to V and 45° 
with H. The base is inscribed in a 3'' circle, the altitude is 
3l/2'^ and the cutting plane is parallel to and 11/2'' above the 
base. 

Prob. 26. Truncated Oblique Pentagonal Pyramid. The 
base is parallel to H with the left side of base perpendicular 
to V. The axis is inclined to the right so that the apex is 214'' 
to the right of the center of the base. The base is inscribed in 
a 2%'' circle, the altitude is 3%", the cutting plane is perpen- 
dicular to V at any desired angle with H, so that the highest 
point is 1%" above the base on the extreme left-hand side. 

Prob 27. A pyramid having its rectangular base, 1^^^ by 
parallel to H and being 3%'' in altitude has the longer 
sides of the base parallel to V. The pyramid is cut by a plane, 
30° with H and perpendicular to V, at a point 1^' above the 
base on the right. 

Prob. 28. Truncated Cylinder. A cylinder with base 
parallel to H is cut by a plane, perpendicular to V, so that the 
cyhnder is 3'' high on the extreme left-hand element and %'' 
high on the extreme right. The cylinder is 21^4'' in diameter. 
Show auxiliary view, profile view and development of curved 
surface. Study notes on the cylinder carefully. Page 12. 



21/4^ 




r/G. 32 

Prob. 29. Make the 'stretch-out' of the sheet metal re- 
turn, the bottom view of which is shown in Fig. 32. Scale, 
full size. 



31 



Prob. 30. Truncated Cone. A right circular cone, with 
base parallel to H and 2%'^ in diameter, is truncated by a 
plane, perpendicular to V and at 60° on the right with H, so 
that the extreme right-hand element is cut %'' above the 
base. The original altitude of the cone was 3%''. Page 13. 



Prob. 31. Truncated Oblique Cone. An oblique cone with 
a circular base, ^V^!^ in diameter, is Zy^'' in altitude. Its apex 
is 2'' to the right of the center of the base and its axis is 
parallel to V, while its base is parallel to H. It is cut by a 
plane perpendicular to V and at an angle of from 45° to 60° 
with H. The cutting plane intersects the right-hand element 



1/2" 



above the base. 



£LL /P'Se: 



PA /^A SOLA 




HYPeRBOLA 



P/G. 33 



Prob. 32. Conic Sections. Taking a right circular cone 
V^ in diameter at the base, 3%^^ in altitude, cut it by planes 
through a point on the right 23/4'' above the base; the first 
parallel to the left-hand element, the second at about 45° to 
H, the third perpendicular to the base. All the planes are to 
be perpendicular to V. Find the revolved sections. Fig. 33. 



32 



Intersections and Developments 

Show all the necessary projections and the May-outs,' or 
developments, complete. See pages 24 and 25. 

Prob. 33. Pipe Tee. Diameter of main 21/2''; of off -set 
1%'\ axes in the same plane. Make the height of the larger 
cylinder 3'' with the axis vertical; and end of off-set 2'' from 
axis of main. 




/^/a v3-f 



Prob. 34. Sheet Metal Pipe Connection. Place the cen- 
ter of the H projection of the intersecting cylinders, indicated 
in Fig. 34, 7l^'' to the right of the left-hand border and 914''* 
above the lower border of a C-size sheet. Place the bottom 
of the V projection 4%'' above the lower border. At the right, 
show the profile view as well as the development of the oblique 
cylinder; at the left, show the profile view and the develop- 
ment of the horizontal cyUnder; along the lower portion of 
the drawing show the development of the vertical cylinder- 
Scale, Q''=r-0'\ 



33 



NOTE: — An examination of the four-piece elbow, Fig. 35, 
will show that the truncated cylinders forming the end pieces 
are cut at a right angle with the axis on one end and at 15° 
to the base on the other. In the two intermediate pieces, the 
joint forms an angle of 15° with the center line of the piece. 
The angle of the elbow, 90°, is divided into six equal angles 
of 15°, formed by the joint between the pieces and the center 
line of the piece. For an elbow of any angle, there are as 
many angles formed as there are end pieces plus twice the 
number of intermediate pieces. 

Example: — In a four-piece elbow there are, 
2 end pieces =2 angles 

2 intermediate pieces =4 angles 



Sum of angles ^6 

90°-^6=15° in each angle. 





r/G. 35 



^/G. 36 



Prob. 35. Four-piece elbow. Fig. 35. Scale, 4''-=l'-0'^ 
Prob. 36. Three-piece elbow, Fig. 36. Scale, 6''=l'-0'^ 



34 



Prob. 37. Five-piece elbow, Fig. 37. Scale, 4''=l'-0' 




rtG. 3 7 



Prob. 38. Grocer's Scoop, Fig. 38. Scale, 4:''=V-0'\ 




6E:CTfON ON A A 



rtG. 38 



35 



Prob. 39. Funnel, Fig. 39. Scale, Q''=V-0'\ 








,J 


I 


1 colt 


\ 




/ 1 


Y'¥ 


1 


V 


/ 1 



^iv 



Prob. 40. Ventilator, Fig. 40. Scale, 2''=V-0'\ Place 
the center of the H projection at 2%''-S%^^, using a C-size 
sheet. The bottom of the cylindrical piece, in the V projection, 
3%'^ above the lower border. In the lay-out of the bottom 
piece, make the seam down the longest side. 




FIG. 40 

36 



Prob. 41. Transition Piece, Fig. 41. Scale, 3''=l'-0''. 
See second method of finding the true length of line, page 10. 




Prob. 42. Transition Piece with off-set, Fig. 42. Scale, 




Prob. 43. Connector, Fig. 43. Scale, full size. 




r/G. ^3 
Prob. 44. Connector, Fig. 44. Scale, full size. 




F/G. -^-^ 

38 



Prob. 45. Reduction elbow, Fig. 45. Scale, Q''=r-0'\ 




r/G. 4^5 



Prob. 46. Oil-Bucket, Pig. 46. Scale, 6'^=l'-0^ 




rtG. -^/^ 



39 



Prob. 47. Intersecting Cone and Square Prism, Fig. 47. 
Scale, full size. 




F/G. -f 7 



Prob. 48. Intersecting Square Pyramid and Square 
Prism, Fig. 48. Scale, full size. 



A 




5 ► 


_L i 




H 3.88'^— A 

r/G. ^6 



40 



Prob. 49. Make the 'lay-out' of the small tower shown in 
Fig. 49. Scale, V'=V-0'\ 




r/G. ^9 



Prob. 50. Find the true size of the dihedral angle formed 
by the sides of the lamp shade shown in Fig. 50. Scale, 




r/G. so 



41 



